1. Instructor: Aaron Roth aaroth@cis.upenn.edu
CIS 262 Automata, Computability, and Complexity Spring
Instructor: Aaron Roth
Lecture: April 17, 2019

2. NP-Complete Problems A problem A is called NP-complete if
1. A is in NP
2. A is NP-hard: for every language L in NP, L <P A
Cook’s Theorem (1972):
SAT is NP-complete
Proof: Need to show that every problem in NP poly-time-reduces to SAT
We showed that COLOR <P SAT
COLOR : Given graph G and k, can G be colored using k colors ?
Need to generalize the proof using poly-time nondeterministic TMs
Will get back to this later…

3. Invalidity
INVALID = {<j> | there exists a truth assignment r s.t. r(j) = 0 }
We know that INVALID is in NP
Show that SAT <P INVALID
Reduction: Algorithm for SAT using algorithm R for INVALID
Given a Boolean formula j,
Check if R accepts ~ j : if so, accept else reject
Since SAT is NP-complete, it follows that INVALID is NP-complete
Claim: If a problem A is in NP and B <P A for a known NP-complete problem B, then A is also NP-complete
Proof : ??

4. NP-Complete Problems NP
Invalid
Coloring
Indep Set
SAT
CNF-SAT
Clique
Hamiltonian Path
3SAT
A large class of problems are all computationally “equivalent’
An efficient algorithm for one can be used to solve others
We have seen some reductions, and will see a few more

5. Formulas in Conjunctive Normal Form (CNF)
Boolean formulas:
Boolean variables x, y, z …
Logical operators: And (&), Not (~), Or (|)
Formula in CNF: Nesting of logical operators is restricted
~ at lowest level, | at mid level, & at top level
Example:
( x | ~y | u ) & ( ~x | y | ~z ) & ( ~u | y | ~z )
Literal : a variable or a negated variable
Clause : OR (also called disjunction) of literals
CNF Formula : AND (also called conjunction) of clauses

6. 3CNF A CNF Boolean formula is a conjunction (&) of clauses,
where each clause is a disjunction (|) of literals,
where each literal is a variable or a negated variable
A CNF formula is in 3CNF if each clause has only three literals
( x | y | ~z ) & (~x | u | w ) & (~u | w | z ) & ( z | ~w | ~y ) : 3CNF
(x | z | ~y | u ) & ( ~x | y | z ) : CNF but not 3CNF
( x & y ) | (~x & ~ y) : not CNF

7. SAT, CNF-SAT, and 3SAT
SAT: Given a Boolean formula j, check if it is satisfiable
CNF-SAT: Given a Boolean formula j in CNF, check if it is satisfiable
3SAT: Given a Boolean formula j in 3CNF, check if it is satisfiable
Claim: SAT, CNF-SAT, 3SAT are in NP
Claim: SAT <P CNF-SAT
Given a boolean formula j, we can construct a formula y in CNF such
that j is satisfiable iff y is satisfiable and |y| is polynomial in |j|
Details of proof: see textbook
Claim CNF-SAT <P 3SAT (later today)
Next: 3SAT <P IndepSet
It follows that SAT <P IndepSet, and IndepSet is NP-complete

8. Satisfying Assignment for 3CNF Formulas
( x | y | ~z ) & (~x | y | w ) & (~y | w | z ) & ( z | ~x | ~y )
To make this formula True, each clause must be made True, so at least one literal from each clause must be made True
Search for satisfying assignment for a formula in 3CNF
Pick one literal from each clause (e.g. either x or y or ~z from clause 1)
Cannot choose conflicting literals, that is, a variable and its negation
(e.g. cannot pick literal x from clause 1 and literal ~x from clause 2)
For above formula, we can pick x, w, w, and ~y
Satisfying assignment: x=1, w=1, y=0 (value of z doesn’t matter)

9. Reducing 3SAT to IndepSet
( x | y | ~z ) & (~x | y | w ) & (~y | w | z ) & ( z | ~x | ~y )
Satisfying assignment: One literal per clause, but no conflicting literals
Goal: Construct a graph G such that independent sets of certain size in that graph correspond to satisfying assignment of formula x ~z y ~x w y ~y z w z ~y ~x
Graph: Node for each literal and edges between literals of same clause
+ Edges connecting conflicting literals in different clauses
Indep set of size 4 in graph  Satisfying assignment to formula

10. Reducing 3SAT to IndepSet
Given a Boolean formula j in 3CNF with k clauses,
construct a graph G with 3k nodes, one per each literal
add edges between nodes corresponding to literals in same clause
add edges between nodes corresponding to conflicting literals
G has an independent set with k nodes iff j is satisfiable
G has k triangles, one per clause, so indep set of size k must include
one node from each triangle.
Chosen nodes cannot correspond to conflicting literals
Thus to satisfy j, just set literal for each chosen node to True
To check if j is satisfiable, check if <G,k> is in IndepSet
Size of G is polynomial in | j |, so 3SAT <P IndepSet

11. Reducing CNF-SAT to 3SAT
CNF-SAT: Given a Boolean formula j in CNF, check if it is satisfiable
Given formula is a conjunction of clauses, but some clauses may not have exactly 3 literals
( x ) & ( x | ~y | z ) & ( x | y | z | w) & ( ~y | z | ~w | u | ~v ) …
Goal: How to replace such clauses with 3-literal clauses without affecting satisfiability ?
Handling clauses with < 3 literals: trivial, just duplicate literals
x is same as x | x | x
x | ~y is same as x | ~y | ~y
So let’s assume each clause has 3 or more literals

12. Reducing CNF-SAT to 3SAT
Consider clause ( x | y | z | w)
Introduce a fresh (i.e. new) variable x’
Intended meaning: x’ should be equivalent to z | w
Replace original clause by (x | y | x’)
Add clauses ( x’ | ~z) , ( x’ | ~w), and ( z | w | ~x’)
(these are constraints for x’  (z | w) and (z | w)  x’ in CNF form)
Claim: In a satisfying assignment for ( x’ | ~z ) & ( x’ | ~w) & ( z | w | ~x’)
either x’ = 1 and either z =1 or w=1
or x’ = 0 and both z= 0 and w =0
Claim: ( x | y | z | w) & j is satisfiable if and only if
(x | y | x’) & ( x’ | ~z ) & ( x’ | ~w) & ( z | w | ~x’) & j is satisfiable

13. Reducing CNF-SAT to 3SAT
Consider clause ( x | … | u | v ) with four or more literals
Introduce a fresh variable x’ to stand for u | v
Replace original clause by (x | … | x’)
Add clauses ( x’ | ~u) , ( x’ | ~v), and ( u | v | ~x’)
New clauses have <= 3 literals
Replacement clause has one fewer literal than the original one
Apply this process repeatedly until no clauses with >3 literals
Each step preserves satisfiability
Number of steps is linear in size of original formula
Final formula is in 3CNF and has size polynomial in original formula
Thus CNF-SAT <P 3SAT

14. 3SAT vs 2SAT
2CNF: Conjunction of clauses, where each clause has 2 literals
2SAT : Given a formula j in 2CNF, is it satisfiable ?
Why can’t we reduce 3SAT to 2SAT using same idea of previous proof?
Introducing a fresh variable x’ to stand for u | v to reduce number of literals in a clause needs a clause with 3 literals ( u | v | ~x’)
2SAT is in P
there is a nice poly-time algorithm to solve 2SAT

15. SAT is NP-complete Cook’s Theorem (1972): SAT is NP-complete Proof:
1. SAT is in NP (easy to show)
2. SAT is NP-hard
Consider a language L in NP
Need to show that L <P SAT
Consider a poly-time verifier V for L
A given string w is in L iff there is some certificate c s.t. V accepts <w,c>
Suppose |w| = n
Suppose time complexity of V is f(n), which must be O(nk)
Let N = f(n)
Goal: construct a formula j such that | j | is polynomial in n and is satisfiable if and only if there exists c such that V accepts <w,c>

16. Computation of Verifier TM
a b a a b # _ _ _ _ q0
V executes for only N steps
V uses only N tape cells
Input to V looks like: Input w # Certificate c , followed by blanks
Question : Given w, is there c such that V accepts c
Desired formula depends on w and V
Suppose TM V = (Q, q0, qa, qr, G, S, _, d )

17. Encoding Computation of V
a b b a b # a # b a a _ _ _ _ q3
TM V = (Q, q0, qa, qr, G, S, _, d )
Uses only N tape cells and executes for N steps
A configuration can be encoded as (u , q, v) with u, v in G* with |u|+|v|=N
So a configuration can be encoded as a vector of length N+1
e.g.: above configuration is [ a b b a q3 a b # a # b a a _ _ _ _]
Hence, the entire execution can be encoded as a (N+1) x (N+1) matrix
C[ i , j ] : element of G U Q
First row of C = Initial configuration of V
i-th row of C = configuration after (i-1) steps
(N+1)-th row of C = Configuration upon termination

18. Checking is w in L ? w is in L iff V accepts w # c for some c
Is there an accepting execution of V on some input of the form w # c ?
Is there a (N+1) x (N+1) matrix C with elements in G U Q such that
(1) First row of C starts with q0 w #
Note: value of certificate c is left unspecified
(2) For i = 2, … (N+1), i-th row encodes successor of configuration encoded by (i-1)-th row according transition function of V
(3) Last row encodes accepting configuration, i.e. state equals qa

19. Encoding using Boolean Variables
Are there variables C[i,j], for i,j=1,…(N+1), ranging over G U Q such that constraints (1), (2), and (3) from previous slide are satisfied
But we have only Boolean variables ??
Suppose elements in G U Q are numbered 1, 2, … H
Boolean variables B[i, j, k], for i,j = 1, …, (N+1), and k = 1, … H
C[i,j] = a is encoded by setting B[i,j,a] = 1 (and B[i,j,a’] = 0 for a !=a’ )
Desired Boolean formula is AND of many formulas including following:
for every i,j=1,…(N+1), there is exactly one a such that B[i,j,a] is True

20. Initialization Constraint
(1) First row of C starts with q0 w #
Note: value of certificate c is left unspecified
B[ 1, 1, q0 ] & B[ 1, 2, w1] & B[ 1, 3, w2] & … & B[1, n+1, wn] & B[1, n+2, #]
Note: remaining values in first row (i=1) are unconstrained
Satisfying assignment will tell us what the certificate is

21. Acceptance Constraint
Is there a (N+1) x (N+1) matrix C with elements in G U Q such that
(3) Last row encodes accepting configuration, i.e. state equals qa
B[ N+1, 1, qa ] | B[ N+1, 2, qa ] | … | B[ N+1, N+1, qa ]

22. Adjacent Rows are Successor Configurations
Is there a (N+1) x (N+1) matrix C with elements in G U Q such that
(2) For i = 2, … (N+1), i-th row encodes successor of configuration encoded by (i-1)-th row according transition function of V
a a b a a b q2 a b … _
a a b a a b a’ q3 b … _ (assuming d(q2, a) = (q3, a’, R)
a a b a a b q4 a’ b’ … _ (assuming d(q3, b) = (q4, b’, L)
Observation: C[i, j] depends only on C[ i-1,j-1], C[i-1, j], C[i-1,j+1], C[i-1,j+2]

23. Adjacent Rows are Successor Configurations
a a b a a b q2 a b … _
a a b a a b a’ q3 b … _ (assuming d(q2, a) = (q3, a’, R)
a a b a a b q4 a’ b’ … _ (assuming d(q3, b) = (q4, b’, L)
Observation: C[i, j] depends only on C[ i-1,j-1], C[i-1, j], C[i-1,j+1], C[i-1,j+2]
Desired formula says that for all i, j, value of C[i,j] is consistent with values C[ i-1,j-1], C[i-1, j], C[i-1,j+1], C[i-1,j+2] and transition function of V
For example: B[i-1,j,q2] & B[i-1,j+1,a]  B[i,j,a’] & B[i,j+1,q3]
Details a bit tedious

24. Completing the proof
Given a verifier V of time complexity f(n) for an NP language L and input w,
construct a formula j with (N+1)*(N+1)*|G U Q| variables
Satisfying assignment for the formula j corresponds to filling in
a (N+1) x (N+1) matrix C with elements in G U Q such that
(1) First row of C starts with q0 w #
(2) For i = 2, … (N+1), i-th row encodes successor of configuration encoded by (i-1)-th row according transition function of V
(3) Last row encodes accepting configuration, i.e. state equals qa
Note: satisfying assignment for variables for the rest of the first row gives the value of the certificate c that makes V accept
Thus to check if w is in L, we can check if j is satisfiable
And this proves Cook’s theorem that SAT is NP-complete